Problem: What is the extraneous solution to these equations? $\dfrac{x^2}{x + 9} = \dfrac{4x + 32}{x + 9}$
Solution: Multiply both sides by $x + 9$ $ \dfrac{x^2}{x + 9} (x + 9) = \dfrac{4x + 32}{x + 9} (x + 9)$ $ x^2 = 4x + 32$ Subtract $4x + 32$ from both sides: $ x^2 - (4x + 32) = 4x + 32 - (4x + 32)$ $ x^2 - 4x - 32 = 0$ Factor the expression: $ (x - 8)(x + 4) = 0$ Therefore $x = 8$ or $x = -4$ The original expression is defined at $x = 8$ and $x = -4$, so there are no extraneous solutions.